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TomLikesPhysics

  • 3 years ago

Where is the gradient of phi a) perpendicular b) parallel to the z-axis?

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  1. TomLikesPhysics
    • 3 years ago
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  2. TomLikesPhysics
    • 3 years ago
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    I guess I have to set something zero for a) and perhaps in b) z should be constant?

  3. myko
    • 3 years ago
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    use dot product. If need more help tell. :)

  4. TomLikesPhysics
    • 3 years ago
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    I do the dot thing to get the parellel part and a cross product for the perpendicular part?

  5. myko
    • 3 years ago
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    yes. to find where it is perpendicular to z axis solve this equation: nabla f . (0,0,1)=0 for paralel : nabla f . (0,0,1)= |nabla f|

  6. myko
    • 3 years ago
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    other way would be finding maximum and minimum of |nabla f|

  7. TomLikesPhysics
    • 3 years ago
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    So for the parallel part I dont use a any vector but the vector (0,0,1) and dot this vector with nabla?

  8. TomLikesPhysics
    • 3 years ago
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    for a) I ended up with z^2=xy Is that my final answert for a set of points for which the gradient is parallel to the z-axis?

  9. TomLikesPhysics
    • 3 years ago
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    for b) I got now y^2-xz=x^2-yz This looks also not very good as a final answer.

  10. myko
    • 3 years ago
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    for b): 3(x^2-yz,y^2-xz,z^2-xy).(0,0,1)=3sqrt((x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2)=> (z^2-xy)=sqrt((x^2-yz)^2+(y^2-xz)^2+(z^2-xy)^2) now should try to simplify this somehow

  11. myko
    • 3 years ago
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    noticing that it will happen when : (x^2-yz)^2=0 and (y^2-xz)^2=0

  12. myko
    • 3 years ago
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    so x^2=yz and y^2=xz from 1º: y=x^2/z putting this in 2º: x^4/z^2=xz => x^3=z^3 =>x=z=y which is a line passing through the origin forming 45º with every axis

  13. myko
    • 3 years ago
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    @TomLikesPhysics

  14. TomLikesPhysics
    • 3 years ago
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    So x, y, and z can be any number but as long as the have all the same value the gradient of phi is parallel to the z-axis?

  15. TomLikesPhysics
    • 3 years ago
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    :( c) When is the gradient zero. There I already calculated that the gradient would be zero for x=y=z so it seems kind of weird that this is the same solution for the gradient to be parallel to the z-axis.

  16. TomLikesPhysics
    • 3 years ago
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    But you calculated x=y=z so how can x=y=0 and z= anything?

  17. myko
    • 3 years ago
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    Ya that's true. Forget it.

  18. myko
    • 3 years ago
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    On the other hand as much as I remmeber vector that is 0 is also considered perpendicular or parallel to anything

  19. TomLikesPhysics
    • 3 years ago
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    This is confusing. Did I at least got a) right? I ended up with z^2=xy but that looks kind of weird for a final answert, but I don´t know if I can simplify or clarify that anymore.

  20. myko
    • 3 years ago
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    Yes it's right. Here you got the graph of that: http://www.wolframalpha.com/input/?i=z%5E2%3Dxy+

  21. myko
    • 3 years ago
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    I think part b) is ok too. Not sure tough, how to explain that it is same as for c)

  22. TomLikesPhysics
    • 3 years ago
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    Hmmmm ... I am still pretty confused but if you think this is alright than I will stick to it. Thank you for your help, especially since this took so long. Thanks a lot, myko.

  23. myko
    • 3 years ago
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    Glad to help. Sry for not clarifying it till the end...

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